The given series is S=3+7+13+21+31+....
⇒S=3+7+13+21+31+...an−1+an
⇒S=3+7+13+21+31+...an−2+an−1+an
On subtracting both the equation, we obtain
S−S=[3+(7+13+21+31+...+an−1+an)]−[(3+7+13+21+31+...+an−1+an]
⇒S−S=3+[(7−3)+(13−7)+(21−13)+...(an+an−1)]an
⇒0=3+[4+6+8+...(n−1)terms]−an
⇒an=3+[4+6+8+...(n−1)terms]
⇒an=3+(n−12)[2×4+(n−1−1)2]
=3+(n−12)[8+(n−2)2]
=3+(n−12)(2n+4)
=3+(n−1)(n+2)
=3+(n2+n−2)
=n2+n+1
∴S=∑nk=1ak=∑nk=1k2+∑nk=1k+∑nk=11
=n(n+1)(2n+1)6+n(n+1)2+n [by using ∑ni=1k2=12+22+32+…+n2=n(n+1)(2n+1)6]
=n[(n+1)(2n+1)+3(n+1)66]
=n[2n2+3n+1+3n+3+66]
=n[2n2+6n+106]
=n3(n2+3n+5)